Đáp án:
$S=\left\{\dfrac{k\pi}{2}\,\bigg{|}\,k\in\mathbb Z\right\}$
Giải thích các bước giải:
$\sin x+\sin3x-2\sin2x=0$
$⇔2\sin2x\cos x-2\sin2x=0$
$⇔2\sin2x(\cos x-1)=0$
$⇔\left[ \begin{array}{l}2\sin2x=0\\\cos x-1=0\end{array} \right.⇔\left[ \begin{array}{l}\sin2x=0\\\cos x=1\end{array} \right.$
$⇔\left[ \begin{array}{l}2x=k\pi\\x=k2\pi\end{array} \right.\,(k\in\mathbb Z)⇔\left[ \begin{array}{l}x=\dfrac{k\pi}{2}\\x=k2\pi\end{array} \right.\,(k\in\mathbb Z)$
$⇒x=\dfrac{k\pi}{2}$
Vậy $S=\left\{\dfrac{k\pi}{2}\,\bigg{|}\,k\in\mathbb Z\right\}$.
